$f(x)$ is such that $f'(x) = \sin \frac{1}{4}x - \cos 4x$. Given that $f(2\pi) = 1$, show that $16f''(x) + f(x) = a \sin 4x + b$, where $a$ and $b$ are constants.

1. Find the second derivative $f''(x)$ by differentiating $f'(x)$.

$f''(x) = \frac{d}{dx} \left( \sin \frac{1}{4}x - \cos 4x \right)$
$f''(x) = \frac{1}{4} \cos \frac{1}{4}x - (-4 \sin 4x)$
$f''(x) = \frac{1}{4} \cos \frac{1}{4}x + 4 \sin 4x$

2. Find the general expression for $f(x)$ by integrating $f'(x)$.

$f(x) = \int \left( \sin \frac{1}{4}x - \cos 4x \right) dx$
$f(x) = \frac{-\cos \frac{1}{4}x}{\frac{1}{4}} - \frac{\sin 4x}{4} + C$
$f(x) = -4 \cos \frac{1}{4}x - \frac{1}{4} \sin 4x + C$

3. Use the given condition $f(2\pi) = 1$ to find the constant of integration $C$.

$f(2\pi) = -4 \cos \left(\frac{1}{4}(2\pi)\right) - \frac{1}{4} \sin (4(2\pi)) + C$
$1 = -4 \cos \frac{\pi}{2} - \frac{1}{4} \sin 8\pi + C$
$1 = -4(0) - \frac{1}{4}(0) + C$
$C = 1$
$f(x) = -4 \cos \frac{1}{4}x - \frac{1}{4} \sin 4x + 1$

4. Evaluate $16f''(x) + f(x)$ to show it is in the required form.

$16f''(x) + f(x) = 16 \left( \frac{1}{4} \cos \frac{1}{4}x + 4 \sin 4x \right) + \left( -4 \cos \frac{1}{4}x - \frac{1}{4} \sin 4x + 1 \right)$
$= 4 \cos \frac{1}{4}x + 64 \sin 4x - 4 \cos \frac{1}{4}x - \frac{1}{4} \sin 4x + 1$
$= \left( 64 - \frac{1}{4} \right) \sin 4x + 1$
$= \frac{255}{4} \sin 4x + 1$
(This is in the form $a \sin 4x + b$ where $a = \frac{255}{4}$ and $b = 1$)